Exercises: systems with two variables

Step 0: Tidying up the equations and selecting a variable

Since the equations are already nicely ordered this point is not needed.

In this case it is the same effort to solve the problem, no matter if you select $x$ or $y$.

The solution is calculated for the variable $x$.

Step 1: Multiply the equations

In equation $I$ there are $4x$, in equation $II$ there are $6x$. So a common multiple is $12x$. Multiply the first equation by $3$ and the second equation by $2$.

Equation 1

$\begin{array}{lcccccll}\mathrm{I}& 4x& +& 2y& =& 4& & |\cdot 3\\ {\mathrm{I}}^{\prime }& 12x& +& 6y& =& 12& & \end{array}$

Equation 2

$\begin{array}{lcccccll}\mathrm{II}& 6x& -& 3y& =& -3& & |\cdot 2\\ {\mathrm{II}}^{\prime }& 12x& -& 6y& =& -6& & \end{array}$

Step 2: Elimination of a variable by addition/subtraction.

The signs of the selected variable are the same both times. Therefore you have to subtract.

$\begin{array}{llcccll}& {\mathrm{I}}^{\prime }& 12x& +& 6y& =& 12\\ -& {\mathrm{II}}^{\prime }& 12x& -& 6y& =& -6\\ & & 0& +& 12y& =& 18\end{array}$

Step 3: Determine the values of both variables.

Determine the value of $y$

$\begin{array}{rrll}12y& =& 18& |:12\\ y& =& {\displaystyle \frac{3}{2}}& \end{array}$

Determine the value of $x$ by plugging $y$ into $\mathrm{II}$

$\begin{array}{rrl}6x-3\cdot {\displaystyle \frac{3}{2}}& =& -3\\ 6x-{\displaystyle \frac{9}{2}}& =& -3& |+{\displaystyle \frac{9}{2}}\\ 6x& =& {\displaystyle \frac{3}{2}}& |:6\\ x& =& {\displaystyle \frac{1}{4}}\end{array}$

So the solution is given by $x={\displaystyle \frac{1}{4}}$ and $y={\displaystyle \frac{3}{2}}$: $𝕃=\{(x|y)=\left({\displaystyle \frac{1}{4}}\right|{\displaystyle \frac{3}{2}})\}$.